CBSE Class 10 Mathematics Areas Related to Circles Notes Set 06

Areas of Combinations of Plane Figures

In our daily life, we come across figures which are combinations of some plane figures. Let us learn how to calculate the areas of combinations of plane figures with the help of some examples.

Example 1. Find the area of the shaded region, where PQRS is a square of side 28 cm.
Answer:

Side of the square \( = 28 \) cm. Area of the shaded region \( = \) area of square PQRS \( - \) area of four quadrants.
Area of square \( PQRS = 28 \times 28 = 784 \) cm². The radius of each quadrant is \( r = \frac{28}{2} = 14 \) cm.
Area of one quadrant \( = \frac{\pi r^2}{4} = \frac{22}{7} \times \frac{14 \times 14}{4} = 154 \) cm², so the area of 4 quadrants \( = 4 \times 154 = 616 \) cm².
Area of the shaded region \( = (784 - 616) \) cm² \( = 168 \) cm².
Hence, the area of the shaded region is 168 cm².

Example 2. Find the area of the shaded region in the given figure.
Answer:

The diameter of the semicircle PRQ is 28 cm, so its radius is \( \frac{28}{2} = 14 \) cm. The diameters of the semicircles PSO and OTQ are each 14 cm, so their radius is 7 cm.
Area of the shaded region \( = \) area of the semicircle of radius 14 cm \( + 2 \times \) (area of the semicircle of radius 7 cm).
\( = \frac{1}{2} \times \pi \times (14)^2 + 2\left[\frac{1}{2} \times \pi \times (7)^2\right] = \pi\left[\frac{1}{2} \times 196 + 49\right] = \frac{22}{7}[98 + 49] = 462 \) cm².
Hence, the area of the shaded region is 462 cm².

Example 3. A paper is in the form of a rectangle PQRS in which \( PQ = 20 \) cm and \( QR = 14 \) cm. A semi-circular portion with QR as diameter is cut off. Find the area of the remaining part.
Answer:

Length of the rectangle \( PQ = 20 \) cm and breadth \( QR = 14 \) cm. Area of rectangle \( PQRS = 20 \times 14 = 280 \) cm².
The diameter of the semicircle is \( QR = 14 \) cm, so its radius is 7 cm. Area of the semicircular portion cut off \( = \frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77 \) cm².
Area of the remaining part \( = (280 - 77) \) cm² \( = 203 \) cm².
Hence, the area of the remaining part is 203 cm².

Example 4. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diameters as shown in the figure. Find the perimeter and area of the shaded region.
Answer:

\( PS = \) diameter of the circle of radius 6 cm \( = 12 \) cm. As PQ, QR and RS are equal, \( PQ = QR = RS = \frac{12}{3} = 4 \) cm and \( QS = QR + RS = 8 \) cm.
Perimeter of the shaded region \( = \) arc of the semicircle of diameter PS \( + \) arc of the semicircle of diameter QS \( + \) arc of the semicircle of diameter PQ \( = (\pi \times 6 + \pi \times 4 + \pi \times 2) \) cm \( = 12\pi \) cm.
Area of the shaded region \( = \) area of the semicircle of diameter PS \( + \) area of the semicircle of diameter PQ \( - \) area of the semicircle of diameter QS.
\( = \frac{1}{2} \times \frac{22}{7} \times 6^2 + \frac{1}{2} \times \frac{22}{7} \times 2^2 - \frac{1}{2} \times \frac{22}{7} \times 4^2 = \frac{1}{2} \times \frac{22}{7}(36 + 4 - 16) = \frac{1}{2} \times \frac{22}{7} \times 24 = \frac{264}{7} \approx 37.71 \) cm².
Hence, the perimeter is \( 12\pi \) cm and the area is \( \frac{264}{7} \) cm² (about 37.71 cm²).

Example 5. In the given figure, find the area of the shaded region. [Use \( \pi = 3.14 \)]
Answer:

The diameter of the circle is the diagonal QS of the rectangle PQRS: \( QS = \sqrt{QR^2 + RS^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \) cm. Hence the radius of the circle is \( r = \frac{10}{2} = 5 \) cm.
Area of the rectangle \( PQRS = 8 \times 6 = 48 \) cm². Area of the circle \( = \pi r^2 = 3.14 \times 5^2 = 78.50 \) cm².
Area of the shaded region \( = \) area of the circle \( - \) area of the rectangle \( PQRS = (78.50 - 48) \) cm² \( = 30.50 \) cm².
Hence, the required area is 30.50 cm².